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Fundamental frequency of a wire wheel spoke

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  • #1
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Homework Statement


The spoke of a wire wheel is 9.5 cm long, 3.5 mm in diameter, and under tension of 2100 N. The wire is made of steel of density 7860 kg/m3. When struck with a metal tool at its center, the spoke rings at its fundamental frequency. What is that frequency?

Homework Equations


λn.fn=v and velocity:v=sqrt(T/μ) , lemda λn=2L/n and L= length of string
fn=n/2L . sqrt(T/μ)
fundamental≡ the first mode ⇒ n=1
μ is the linear mass and must be in kg/m so we must multiply our density by the area which is π.r^2

The Attempt at a Solution


μ=7860*π*(3.5*10^-3)^2=0.302 kg/m
f=1/(2*9.5*10^-2) . sqrt(2100/0.302)
f= 438.8 Hz
but the answer is incorrect why??
 

Answers and Replies

  • #2
gneill
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I've changed your thread title to more accurately describe the particular problem. Please choose thread titles that are not too general.

μ=7860*π*(3.5*10^-3)^2=0.302 kg/m
Were you given the radius of the spoke?
 
  • #3
79
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Were you given the radius of the spoke?
Ah okay. Thanks very much. I used the diameter as a radius wrongly. :)
Now it will be 877Hz which is the correct answer
 

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